Solve the equation 21x^{2} - 28x + 10 = 0. | vedclass

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Question

(A) The given quadratic equation is $21x^{2} - 28x + 10 = 0$.Comparing this with $ax^{2} + bx + c = 0$,we get $a = 21$,$b = -28$,and $c = 10$.The disc

Vedclass · Accepted Answer

$\frac{2}{3} \pm \frac{\sqrt{14}}{21}i$

Vedclass · Answer

(A) The given quadratic equation is $21x^{2} - 28x + 10 = 0$.Comparing this with $ax^{2} + bx + c = 0$,we get $a = 21$,$b = -28$,and $c = 10$.The discriminant $D$ is given by $D = b^{2} - 4ac$.$D = (-28)^{2} - 4(21)(10) = 784 - 840 = -56$.The roots are given by $x = \frac{-b \pm \sqrt{D}}{2a}$.$x = \frac{-(-28) \pm \sqrt{-56}}{2(21)} = \frac{28 \pm \sqrt{56}i}{42}$.Since $\sqrt{56} = \sqrt{4 	imes 14} = 2\sqrt{14}$,we have:$x = \frac{28 \pm 2\sqrt{14}i}{42} = \frac{28}{42} \pm \frac{2\sqrt{14}}{42}i = \frac{2}{3} \pm \frac{\sqrt{14}}{21}i$.

Solve the equation $21x^{2} - 28x + 10 = 0$.

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